# If f is differentiable on a connected open set S..

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• Sep 22nd 2010, 05:49 PM
Pinkk
If f is differentiable on a connected open set S..
and $\partial_{x_{1}}f(x) = 0$ for all $x\in S$, must $f$ be independent of $x_{1}$, that is, given $a,b$ where $a_{j} = b_{j}$ for all $j \ne 1$, then $f(a)=f(b)$.

So I know how to show this is true if $S$ is convex (because then I can use the Mean Value Theorem), but I am having a hard time coming up with such a function, say on a non-convex subset of $\mathbb{R}^{2}$, where this does not hold. Any hints or ideas would be appreciated, thanks.
• Sep 23rd 2010, 04:49 AM
Opalg
Quote:

Originally Posted by Pinkk
and $\partial_{x_{1}}f(x) = 0$ for all $x\in S$, must $f$ be independent of $x_{1}$, that is, given $a,b$ where $a_{j} = b_{j}$ for all $j \ne 1$, then $f(a)=f(b)$.

So I know how to show this is true if $S$ is convex (because then I can use the Mean Value Theorem), but I am having a hard time coming up with such a function, say on a non-convex subset of $\mathbb{R}^{2}$, where this does not hold. Any hints or ideas would be appreciated, thanks.

An open connected set in $\mathbb{R}^n$ is path-connected. Given two points with a continuous path connecting them, you can split the path into segments each of which lies in a convex subset of S (e.g. an open ball contained in S). So f will be independent of $x_1$ on each segment, and therefore along the whole path.
• Sep 23rd 2010, 05:26 AM
Pinkk
Hmm, that makes sense, but my textbook (without actually showing the example) says you can find such an example if you have an open connected set in $\mathbb{R}^{2}$ with a hole in it. But if you can cover an open connected set with overlapping convex sets, then how can such a counterexample exist?
• Sep 23rd 2010, 07:17 AM
Opalg
Quote:

Originally Posted by Pinkk
Hmm, that makes sense, but my textbook (without actually showing the example) says you can find such an example if you have an open connected set in $\mathbb{R}^{2}$ with a hole in it. But if you can cover an open connected set with overlapping convex sets, then how can such a counterexample exist?

Cancel my previous comment. Here's an example of what the book claims (and it doesn't even need a hole in the domain, just a slit). Let Y be the negative y-axis together with the origin, $Y = \{(0,y):y\leqslant0\}$, and let S be the complement of Y, $S = \mathbb{R}^2\setminus Y$. Define f on S by

$f(x,y) = \begin{cases}y^3&(\text{if }x\geqslant0),\\ |y^3|&(\text{if }x<0).\end{cases}$

If y>0 then $f(x,y) = y^3$ regardless of whether x is positive or negative. So $\frac{\partial f}{\partial x} = 0$ for all x (and f is independent of x).

If y<0 then $\frac{\partial f}{\partial x} = 0$ for x<0 and also for x>0. But f(x,y) will be $y^3$ if x is positive, and $-y^3$ if x is negative. So it depends on x.

Interesting result! (Surprised)
• Sep 23rd 2010, 12:23 PM
zzzoak
If we define sign(x)
$
sign(x) = \begin{cases}1&(\text{if }x\geqslant0),\\ -1&(\text{if }x<0).\end{cases}
$

then
$
f(x,y)=g(y) \; sign(x)
$

and
$
\frac{\partial f}{\partial x} = g(y) \; 2 \delta (x)
$

and
f(x,y) depends on x.